Find the product of all real values of $r$ for which $\frac{1}{2x}=\frac{r-x}{7}$ has exactly one real solution.
Answer: Observe first that $x=0$ is not a solution to the equation since it makes the denominator of $\frac{1}{2x}$ equal to 0.  For $x\neq 0$, we may multiply both sides by both denominators and move all the resulting terms to the left-hand side to get $2x^2-2rx+7=0$. Observe that there are two ways the original equation could have exactly one solution.  Either $2x^2-2rx+7=0$ has two solutions and one of them is 0, or else $2x^2-2rx+7=0$ has exactly one nonzero solution.  By trying $x=0$, we rule out the first possibility.

Considering the expression $\frac{-b\pm \sqrt{b^2-4ac}}{2a}$ for the solutions of $ax^2+bx+c=0$, we find that there is exactly one solution if and only if the discriminant $b^2-4ac$ is zero.  Setting $(-2r)^2-4(2)(7)$ equal to 0 gives $4r^2-4(14) = 0$.  Add 4(14) and divide by 4 to find $r^2=14$.  The two solutions of this equation are $\sqrt{14}$ and $-\sqrt{14}$, and their product is $\boxed{-14}$.